Identify the metal ions and geometries

• In $[Fe(H_2O)_6]Cl_2$, the complex ion is $[Fe(H_2O)_6]^{2+}$. Since $H_2O$ is neutral, oxidation state of Fe is: $x+6(0)=+2 \Rightarrow x=+2$ So metal ion is $Fe^{2+}$.

• In $K_2[NiCl_4]$, the complex ion is $[NiCl_4]^{2-}$. Let oxidation state of Ni be $x$: $x+4(-1)=-2 \Rightarrow x=+2$ So metal ion is $Ni^{2+}$.

Find the $d$-electron configurations

• $Fe$: $[Ar]3d^64s^2$ so $Fe^{2+}$ is $3d^6$.
• $Ni$: $[Ar]3d^84s^2$ so $Ni^{2+}$ is $3d^8$.

CFSE of $[Fe(H_2O)_6]^{2+}$

This is an octahedral complex.

Since $H_2O$ is a weak field ligand, $Fe^{2+}$ ($d^6$) will be high spin: $t_{2g}^4 e_g^2$

CFSE for octahedral field:

• each $t_{2g}$ electron contributes $-0.4\Delta_o$
• each $e_g$ electron contributes $+0.6\Delta_o$

Therefore, $\text{CFSE} = 4(-0.4\Delta_o)+2(+0.6\Delta_o)$ $= -1.6\Delta_o+1.2\Delta_o$ $= -0.4\Delta_o$

CFSE of $[NiCl_4]^{2-}$

This is a tetrahedral complex because $Cl^-$ is weak field and four-coordinate $Ni^{2+}$ with chloride is tetrahedral.

For tetrahedral splitting:

• lower set: $e$
• upper set: $t_2$

For $Ni^{2+}$, $d^8$ configuration in tetrahedral field becomes: $e^4 t_2^4$

CFSE in tetrahedral field:

• each $e$ electron contributes $-0.6\Delta_t$
• each $t_2$ electron contributes $+0.4\Delta_t$

Hence, $\text{CFSE} = 4(-0.6\Delta_t)+4(+0.4\Delta_t)$ $= -2.4\Delta_t+1.6\Delta_t$ $= -0.8\Delta_t$

Match with options

We obtained:

• $[Fe(H_2O)_6]Cl_2 : -0.4\Delta_o$
• $K_2[NiCl_4] : -0.8\Delta_t$

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So, they agree.